Question

Match the two columns;

$\begin{array}{cc}Column-I& Column-II\\ \left(A\right)CisapointontheminorarcABofthecirclewithcentreO.& \left(P\right){60}^{\circ }\\ Given\angle ACB=x.Calculate\angle xifACBOisaparallelogram.& \\ \left(B\right)ChordEDisparalleltothediameterACofthecircle.& \left(q\right){120}^{\circ }\\ Given\angle CBE={55}^{\circ },calculate\angle DEC.& \end{array}$

$\begin{array}{cc}\left(C\right)Inthegivenfigure,Oisthecentreofthecircle.& \left(r\right){35}^{\circ }\\ If\angle ACB={60}^{\circ }find\angle OAB.& \end{array}$



$$\begin{array}{cc}\left(D\right)Inthegivenfigure,Oisthecentreofacircle.& \left(s\right){30}^{\circ }\\ \angle AOB=400and\angle BDC=1000.Find\angle OBC& .\end{array}$$

Choose the Correct option;

• A. A-s; B-p; C-q; D-r
• B. A-r; B-s; C-p; D-q
• C. A-q; B-p; C-r; D-s
• D. A-q; B-r; C-s; D-p

Answer 1

The correct option is D A-q; B-r; C-s; D-p

(d) $\left(A\right)\to q;\left(B\right)\to r;\left(C\right)\to s;\left(D\right)\to p$
(A) Clearly, major arc BA subtends $anglex$ at a point on the remaining part of the circle.
$\therefore$ Reflex $\angle AOB=2x$


$\Rightarrow {360}^{\circ }-y=2x$
$\Rightarrow y={360}^{\circ }-2x$
Thus, $y={360}^{\circ }-2x$
Since ACBO is a parallelogram.
$\therefore x=y\Rightarrow x={360}^{\circ }-2x$
$\Rightarrow 3x={360}^{\circ }\Rightarrow x={120}^{\circ }$

(B) Consider the arc $CDE.\angle CBEand\angle CAE$
Are the angles in the same segment of arc CDE.
$\therefore \angle CAE=\angle CBE$
$\Rightarrow \angle CAE={55}^{\circ }[\angle CBE={55}^{\circ }]$
Since AC is the diameter of the circle and the angle in a semi - circle is a right angle
Therefore,
$\angle AEC={90}^{\circ }$. Now, in $\Delta ACE$, we have
$\angle ACE+\angle AEC+\angle CAE={180}^{\circ }$
$\Rightarrow \angle ACE+{90}^{\circ }+{55}^{\circ }={180}^{\circ }$
$\Rightarrow \angle ACE={35}^{\circ }$.


But $\angle DEC$ and $\angle ACE$ are alternate angles, because $AC||DE$.
$\therefore \angle DEC=\angle ACE={35}^{\circ }$.

(C) $\angle AOB={120}^{\circ }$
(The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)
OA = OB (Radius of the circle)



$x+x+{120}^{\circ }={180}^{\circ }$
$\Rightarrow 2x=60$
$\Rightarrow x=30$

(D) $\angle AOB=2\times \angle ACB$
$\Rightarrow \angle ACB=\frac{1}{2}\times \angle AOB$
$=\frac{1}{2}\times {40}^{\circ }={20}^{\circ }$



In $\Delta BDC,\angle DBC+\angle BDC+\angle DCB={180}^{\circ }$
$\Rightarrow \angle OBC+{100}^{\circ }+{20}^{\circ }={180}^{\circ }$
$\therefore \angle OBC={180}^{\circ }-{120}^{\circ }={60}^{\circ }$